A particle moving with a circle of radius 40 cm has a linear speed of 30m/s at an instant its speed is increasing at the rate of 4m/s². then the rate of change of centripetal acceleration increasing at the instant will be:
a) 200m/s³ b) 600 m/s3 c) 100 m/s² d) 300 ​m/s3

Dear Student,

Please find below the solution to the asked query:

Equation for the centripetal acceleration is,

$a_r=\frac{V^2}{r}$
Where r = radius and V = velocity,

Given information is, at an instant

$r=40 cm=0.4 m; V=30 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.; a=\frac{dV}{dt}=4 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$
So,

$$\begin{gathered} a_r=\frac{V^2}{r} \Rightarrow \frac{d{a}_r}{dt}=\frac{2V}{r}\frac{dV}{dt} \Rightarrow \frac{d{a}_r}{dt}=\frac{2\times 30}{0.4}\times 4 \Rightarrow \frac{d{a}_r}{dt}=20\times 30 \\ \Rightarrow \frac{d{a}_r}{dt}=600 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^3$}\right. \end{gathered}$$

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